## Physics Of Light

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• ### Junaid I

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A Presentation On the Physics of Light Starting With Reflection, Refraction, Diffraction.

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Workshop 3 Workshop 3 Dr Laila. K. Habib Junaid .1.Ziaee
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1- The wavelengths of light visible to the human eye are about 350 nm to 750 nm. What are the corresponding frequencies and energies ? = 8.57 x 101% lor 8.57 x 1014Hz Solution: The corresponding frequencies and energies for : a) 350 nm Wavelength : 1 C (3 x 108)m/s (350 x 10-9)m E = [(6.626 x 10-34)Js x (8.57 x 1014)s- Thus,E = 5.6 x 10 19] b) 750 nm Wavelength : c E=hf (3 x 108)m/s = 4x 1014s 1014Hz (750 x 10-9)m = [(6.626 x 10 34)Js x (4 x 1014)s-1] Thus,E = 2.65 x 10 19]
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Photon A Red Green InfraRed Visual X-rays avelength lar er the same smaller larger th smaller larger the same smaller larger the same smc Iler larger the same smaller Frequency larger the same smaller larger the same smaller larger the same S r arger the same smaller larger the ne smaller Energy larger the same smaller larger the same smaller larger the same smaller arger t e same smaller larger th smaller Velocit) in s ace larger the sam sma er he sa smaller e sa smaller larger e same er he same sma ler Photon B Blue Orange Visual Microwave Gamma-ray 2- Select the correct relationships Wavelength (in meters
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X-Rays Ix 10 12 1 x 10 14 Wavelength (in meters) 4 x 107 High Energy Gamma Rays 1 x 104 Ultraviolet Rays Ix108 5 x 107 Infrared Rays Visible Light 6 x 107 Radar FM I x 102 TV Shortwave AM I x 102 Low Energy I x 104 Wavelength (in meters)
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3- What is the angle of incidence on an air-to-glass boundary if the angle of refraction on the glass (n = 1.52) is 25 degrees? a) 16 degrees b)25 degrees degrees d)43 degrees Solution: We use Snell's law: nt = 1.00(air) n r = 1.52(glass) Or = 250 (Angle on glass) or niSin91= n2Sin92 Ix Sin9i = Sin250 1.52 x Sin250 Sin9i= 1 o'. Or = sin-1(.642) = 39.96 400 Answer : c) niSin9i = nrSin0 = (0642)
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4- When yellow light of wavelength 580 nm passes from air (n =1.00) into water (n=l .33), how does each of the following change, and what are their values in water? a) Speed of light. b) Wavelength of light. c) Frequency of light. Solution: When yellow light of wavelength 580 nm passes from air (n =1.00) into water (n=l .33), the speed of light decreases, the wavelength of light decreases, but the frequency does not change. a) Speed of light : 3x108m/s C n - 225563609m/ s 1.33
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Method 1 : c) Frequency of light : 3x108m/s 580 x 10-9m -1 Method 2 : b) Wavelength of light : 1.33 1 Sin 01 Sin 02 12 .12 12 580 x 10-9m 1.33 b) Wavelength of light : m 12 = 436 c) Frequency of light : 436 x 10—9m -1 — 436 nm
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Norma 600 450 Interface 30 150 Air Water
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5- If the critical angle for internal reflection inside a certain transparent material is found to be 48.0 degrees, what is the index of refraction of the material? (Air is outside the material). =1.35 b) c) d) Illustration 1.35 I .48 1.49 0.743 Solution: The critical angle is the angle of incidence (which is always in the more dense material) for which the angle of refraction is 90 degrees. niSin9i = nrSin9r Apply Snell's law equation: niSin480 = Sin900 n I .743 = I 1 .743 or n1Sin01= n2Sin02 Answer : a)
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480 640 820 Interface Medium 1 Material 350 Normal Normal 450 900 600 Air n 2 Medium 2
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(ii) Focal Length (f) (iii) 6- Rank the converging lenses in increasing power (from least powerful to most): L/ (i), (ii), (iii) b) (iii), (ii), (i) c) (ii), (i), (iii) d) (iii), (i). (ii) Solution: P = l/f (inversely proportionality between power and focal length) The focal length decreases from (i) to (iii), So the power must increase The most powerful Lens (Ill) bends the light to a greater degree than the weakest (l) Answer : a)
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Converging (Convex or Positive) Lens 1 2 3 Image Distance Magnification Focal Length Power Uses Real Image Inverted Other side Smaller Positive Negative Positive Positive Glasses Virtual Image Upright Same side Bigger Negative Positive Negative Negative Magnifying Lenses Diverging lens (Concave or Negative) Lens Virtual Image Upright Same side Bigger Negative Positive Negative Negative Glasses The image appears smaller than object (M< 1), is virtual (di negative), and upright (all virtual images are upright)
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7- An object is placed 0.25 m away from lens. The lens forms an image that is 0.167 m away from the lens, upright, and on the same side of the lens as the object. What is the focal length of the lens? a) 0.11 m b) -0.11 m c) 0.50 m -0.50 m Solution: 1 Using the thin lens equation: 1 1 1 1 1 .25m 0.167m f 1 — —2.0m 1 f = -0.50m Is the lens convex or concave? Concave because focal point is negative (sign convention) Note: the sign convention must be used correctly to reach this answer, since it was stated that the image forms on the same side of the lens as the object, the image must be virtual and di must be negative (-0.167m) Answer : d)
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8- A 14 mm tall object is 4.0 mm from a converging lens. If the image is 4.0 mm tall, how far is it from the lens? Describe the image di — —1. Imm b) c) d) Solution: 1.1 mm 14 mm 1.4 mm 8.7 mm image dis tan ce object dis tan ce 4.0cm image height object height image height object height image dis tan ce object dis tan ce 4.0 14 = 0.28 0.28 Note : The image appears smaller than object (M< 1), is virtual (di negative), and upright (all virtual images are upright) Answer : a)
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Visible Light Spectrum Range Violet 4000 AO 400 nm Indigo 4250 AO 425 nm Blue 4700 AO 470 nm 4,000 400 nm 4,250 Å 5500 AO 550 nm Yellow 6000 AO 600 nm Orange 6300 AO 630 nm Red 6650 AO 665 nm Inm = 10 Blue Aqua Yellow Orange Red 425 nm , 700 470 nm ,900 5,500 6,000 Å 6,300 Å 6,650 Å 7,000 Å 700 nm

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